Let the function, $$f(x)=\begin{cases}-3ax^{2}-2, & x < 1\\a^{2}+bx, & x \geq 0\end{cases}$$ be differentiable for all  $$x \in R,  $$ where $$ a>1, b \in R$$.  If the area of the region enclosed by $$ y=f(x) \text{and the line } y= -20 \text{ is } \alpha+\beta\sqrt{3},\alpha, \beta \in Z$$,  then the value of  $$\alpha + \beta \text{ is } $$_______

To solve this, we find $$a$$ and $$b$$ for differentiability, then integrate to find the area.

For $$f(x)$$ to be differentiable at $$x=1$$, it must be continuous and have equal one-sided derivatives.

• Continuity: $$-3a(1)^2 - 2 = a^2 + b(1) \implies b = -a^2 - 3a - 2$$

• Differentiability: $$-6a(1) = b \implies b = -6a$$

Equating both: $$-6a = -a^2 - 3a - 2 \implies a^2 - 3a + 2 = 0$$.

Given $$a > 1$$, we get $$a = 2$$ and $$b = -12$$.

Intersection with $$y = -20$$

• $$-6x^2 - 2 = -20 \implies x^2 = 3 \implies x = -\sqrt{3}$$ (since $$x < 1$$)

• $$4 - 12x = -20 \implies 12x = 24 \implies x = 2$$

The area is $$\int [f(x) - (-20)] \, dx$$:

1. Left part ($$-\sqrt{3}$$ to $$1$$):

$$\int_{-\sqrt{3}}^{1} (18 - 6x^2) dx = [18x - 2x^3]_{-\sqrt{3}}^{1} = 16 + 12\sqrt{3}$$

2. Right part ($$1$$ to $$2$$):

$$\int_{1}^{2} (24 - 12x) dx = [24x - 6x^2]_{1}^{2} = 6$$

Total Area: $$(16 + 12\sqrt{3}) + 6 = 22 + 12\sqrt{3}$$

Comparing to $$\alpha + \beta\sqrt{3}$$:

$$\alpha = 22, \beta = 12 \implies \alpha + \beta = \mathbf{34}$$